The folutions of the cafes of right-angled plane triangles. Sought Given The hyp. One leg I AC and BC the angles The hyp. The an2 AC and gles one leg BC The hyp. The other 3 AC and leg AB one leg BC The an- The hyp. AC 4 gles and one leg BC Solution. As radius is to the sine of A, fo is the hyp. AC to the leg BC (by Theor. I.) : As AC BC :: radius: fin. A (Theor. I.) whofe complement is the angle C. Let the angles be found, by Cafe 2. and then the required leg AB, by Cafe 1. As fine A: radius :: the leg BC to the hyp AC (Theor. I.) The an- The other As fine A: BC :: fine C gles and leg AB: AB (by Theor. III.) Or, 5 one leg BC The two The an 6 legs AB and BC gles as radius: tang. C:: BC AB (by Theor. II.) As AB : BC :: radius : tang. A (by Theorem II.) whofe complement is the angle C. The two The hyp. Let the angles be found, 7 legs AB and BC AC by Cafe 6. and then the hyp. AC, by Cafe 4. A B A B G C D A D G a The folution of the cafes of oblique plane triangles. Given Sought Solution. The angles Either of As fine C: AB :: fine A I and one fide the other BC (by Theor. III.). 2 3 4 AB Two fides an ang. C op. fides BC The other As AB : fin. C :: BC : fin. A angles A (by Theor. IIl.) which added to and ABCC, and the fum fubtracted from 180 gives the other angle ABC, Two fides The other Let the angle ABC be found, AB, BC and fide AC by the preceding cafe, and then it will be, fin. C: AB :: fin. ABC: AC (by Theor. III.) an opp. an gle C Two fides The other As sum of AB and AC: their Two fides The other Let the angles be found by 5 AC, AB and fide BC the laft cafe, and then BC, the incl.ZA. by Cafe 1. All the three An angle, Let fall a perp. BD opp. to the fuppofe Areq. angle: then (by Theor. IV.) fides 6 as AC: fum of AB and BC: their dif.: dift. DG of the perp. from the middle of the bafe; whence, AD being also known, the angle A will be found by Cafe 2. of right-angles. Note, Note, The 2d and 3d cafes are ambiguous, or admit of two different answers each, when the fide AB oppofite the given angle C (fee fig. 2.) is lefs than the given fide BC, adjacent to it (except the angle found is exactly a right one) for then another right-line Ba, equal to BA, may be drawn from B to a point in the base, somewhere between C and the perpendicular BD, and therefore the angle found by the proportion AB (aB): fin. C :: BC: fin. A (or of CaB,) may, it is evident, be either the acute angle A, or the obtufe one CaB (which is its fupplement), the fines of both being exactly the fame. Having laid down the method of refolving the different cafes of plane triangles, by a table of fines and tangents; I fhall here fhew the manner of constructing fuch a table (as the foundation upon which the whole doctrine is grounded); in order to which, it will be requifite to premise the following propofitions. PROPOSITION I. The fine of an arch being given, to find its cofine, verfed fine, tangent, co-tangent, fecant, and co-fecant. IT Let AE be the proposed arch, EF its fine, CF its co-fine, AF its verfed fine, AT its tangent, CT its fecant, DH its co-tangent, and CH its co-fecant. Then (by 8. 2.) we have CF the fquare root of CE2 EF; whence, not A only the co-fine CF, but alfo the verfed fine AF, will will be known. Then because of the fimilar triangles CFE, CAT, and CDH, it will be (by 14. 4.) 1. CF: FE:: CA: AT; whence the tangent is known, 2. CF: CE (CA) :: CA: CT; whence the fecant is known. 3. EF: CF:: CD: DH; whence the co-tangent is known. 4. EF: EC (CD) :: CD: CH; whence the co-fecant is known. 1. That the tangent is a fourth proportional to the co-fine, the fine, and radius. 2. That the fecant is a third-proportional to the co-fine and radius. 3. That the co-tangent is a fourth-proportional to the fine, co-fine, and radius. 4. And that the co-fecant is a third-proportional to the fine and radius. 5. It appears moreover (because AT: AC:: CD (AC): DH), that the rectangle of the tangent and co-tangent is equal to the fquare of the radius (by 10. 4.): whence it likewife follows, that the tangent of half a right-angle is equal to the radius; and that the co-tangents of any two different arches (reprefented by P and Q are to one another, inversely as the tangents of the fame arches: for, fince tang. P x co-tang. P fqu. rad. = tang. Q× co-tang. Q; therefore will co tang, tang. P: co-tang. Q:: tang. Q: tang. P; or as co-tang, P: tang. Q: co-tang. Q: tang. P (by IO. 4.) PROP. II. If there be three equidifferent arches AB, AC, AD, it will be, as radius is to the co-fine of their common difference BC, or CD, fo is the fine CF, of the mean, to half the fum of the fines BE + DG, of the two extremes: and, as radius to the fine of the common difference, fo is the co-fine FO of the mean, to half the difference of the fines of the two ex meeting DG in H and v. For let BD be drawn, interfecting the radius OC in m; alfo draw mn parallel to CF, meeting AO in ʼn ; and BH and mv, parallel to AO, Then, the arches BC and CD being equal to each other (by hypothefis), OC is not only perpendicular to the chord BD, but also bifects it (by 1. 3.) and therefore Bm (or Dm) will be the fine of BC (or DC), and Om its co-fine: moreover mn, being an arithmetical mean between the fines BE, DG of the two extremes (because Bm Dm) is therefore equal to half their fum, and Do equal to half their difference. But, because of the fimilar triangles OCF, Omn and Dʊm, It will be SOC: Om :: CF: mn } OC: Dm :: FO: Dv2 E. D. COROL |